General results

Nebe, G.

doi:10.1107/97809553602060000541

International
Tables for
Crystallography
Volume A1
Symmetry relations between space groups
Edited by Hans Wondratschek and Ulrich Müller

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International Tables for Crystallography (2006). Vol. A1. ch. 1.5, p. 38 | 1 | 2 |

Section 1.5.6.1. General results

Gabriele Nebe^a ^*

^a Abteilung Reine Mathematik, Universität Ulm, D-89069 Ulm, Germany
Correspondence e-mail: nebe@mathematik.uni-ulm.de

1.5.6.1. General results

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The first very easy but useful remark applies to general groups $[{\cal G}]$ :

Remark . Let $[{\cal M} \leq {\cal G}]$ be a maximal subgroup of $[{\cal G}]$ of finite index $[i: = [{\cal G}: {\cal M}] \,\lt\, \infty ]$ . Then $[{\cal M} \leq {\cal N}_{{\cal G}}({\cal M}) \leq {\cal G} ]$ . Hence the maximality of $[{\cal M}]$ implies that either $[ {\cal N}_{{\cal G}}({\cal M}) ={\cal G} ]$ and $[{\cal M}]$ is a normal subgroup of $[{\cal G}]$ or $[ {\cal N}_{{\cal G}}({\cal M}) ={\cal M} ]$ and $[{\cal G}]$ has i maximal subgroups that are conjugate to $[{\cal M}]$ .

The smallest possible index of a proper subgroup is 2. It is well known and easy to see that subgroups of index 2 are normal subgroups:

Proposition 1.5.6.1.1. Let $[{\cal G}]$ be a group and $[{\cal M}\leq {\cal G}]$ a subgroup of index $[2=|{\cal G}/{\cal M}|]$ . Then $[{\cal M}]$ is a normal subgroup of $[{\cal G}]$ .

Proof . Choose an element $[{\sf g}\in {\cal G}]$ , $[{\sf g}\not \in {\cal M}]$ . Then $[{\cal G} = {\cal M} \cup {\sf g}{\cal M} = {\cal M}\cup {\cal M}{\sf g}]$ . Hence $[{\sf g}{\cal M} = {\cal M}{\sf g}]$ and therefore $[{\sf g}{\cal M}{\sf g}^{-1} = {\cal M}]$ . Since this is also true if $[{\sf g} \in {\cal M}]$ , the proposition follows. QED

Let $[{\cal M}]$ be a subgroup of a group $[{\cal G}]$ of index 2. Then $[{\cal M} ]$ $[{\underline{\triangleleft}}]$ $[{\cal G}]$ is a normal subgroup and the factor group $[{\cal G}/{\cal M}]$ is a group of order 2. Since groups of order 2 are Abelian, it follows that the derived subgroup $[{\cal G}_{1}]$ of $[{\cal G}]$ (cf. Definition 1.5.5.2.1) (which is the smallest normal subgroup of $[{\cal G}]$ such that the factor group is Abelian) is contained in $[{\cal M}]$ . Hence all maximal subgroups of index 2 in $[{\cal G}]$ contain $[{\cal G}_{1}]$ . If one defines $[{\cal N}:= \cap \{ {\cal M} \leq {\cal G} \mid [{\cal G}:{\cal M}] = 2 \}]$ , then $[{\cal G}/{\cal N}]$ is an elementary Abelian 2-group and hence a vector space over the field with two elements. The maximal subgroups of $[{\cal G}/{\cal N}]$ are the maximal subspaces of this vector space, hence their number is [2^a-1] , where $[a: = \dim _{{\bb Z}/2{\bb Z}}({\cal G}/{\cal N})]$ .

This shows the following:

Corollary 1.5.6.1.2. The number of subgroups of $[{\cal G}]$ of index 2 is of the form $[2^{a}-1]$ for some $[a \geq 0]$ .

Dealing with subgroups of index 3, one has the following:

Proposition 1.5.6.1.3. Let $[{\cal U}]$ be a subgroup of the group $[{\cal G}]$ with $[[{\cal G}:{\cal U}] = 3]$ . Then $[{\cal U}]$ is either a normal subgroup of $[{\cal G}]$ or $[{\cal G}/{\rm core}({\cal U}) \cong {\cal S}_{3}]$ and there are three subgroups of $[{\cal G}]$ conjugate to $[{\cal U}]$ .

Proof . $[{\cal G}/\,{\rm core}\,({\cal U})]$ is isomorphic to a subgroup of $[{\cal S}ym_{3}]$ that acts primitively on $[\{1,2,3\}]$ . Hence either $[{\cal G}/\,{\rm core}\,({\cal U}) \cong {\cal C}yc_{3}]$ and $[{\cal U}= \,{\rm core}\,({\cal U})]$ is a normal subgroup of $[{\cal G}]$ or $[{\cal G}/\,{\rm core}\,({\cal U}) \cong {\cal S}ym_{3}]$ , $[{\cal U}/\,{\rm core}\,({\cal U}) \cong {\cal C}yc_{2}]$ and there are three subgroups of $[{\cal G}]$ conjugate to $[{\cal U}]$ . QED

References

International Tables for Crystallography (2006). Vol. A1. ch. 1.5, p. 38